我如何从每个单独父节点子节点?节点

2023-09-08 13:13:27 作者:姐不疯只是拽つ

我有一些数据生成XML。

I have some Data Xml..

<main>
  <TabNavigator x="27" y="11" width="455" height="376" id="gh" backgroundColor="#A4B6E9">
    <NavigatorContent width="100%" height="100%" label="Client" id="clientTab"></NavigatorContent>
    <NavigatorContent width="100%" height="100%" label="Admin" id="adminTab"></NavigatorContent></TabNavigator>
    <TitleWindow x="521" y="84" width="377" height="234">
      <DataGrid x="0" y="0" width="375" height="163" borderVisible="true" id="details">
        <columns>
          <ArrayList>
            <GridColumn dataField="Name" id="arrayName"/><GridColumn dataField="Address" headerText="Address"/>
            <GridColumn dataField="Phone_Number" headerText="Phone_Number"/>
          </ArrayList>
        </columns>
      </DataGrid>
      <Button x="139" y="167" height="28" label="Export"/>
    </TitleWindow>
</main>

我用下面的code,用于检索给定的XML的孩子的名字。

I use following code for retrieving the child names of given XML..

private function urlLdr_complete(event:Event):void{
var xmlData:XML=new XML(URLLoader(event.currentTarget).data);                       
for each (var t:XML in xmlData.children()) 
{
   Alert.show(t.Name);
}

不过,我只得到2个孩子(的TabNavigator和titleWindow正在)。如何才能获得在每个父节点的其他孩子?我想单独为孩子每个父母。我怎样才能得到它?任何人都可以帮我吗..?

But I only get 2 children(TabNavigator and TitleWindow).How do I get the other children in each parent node? I want seperate children for each parent. How can I get it? Could anyone help me please..?

推荐答案

您需要使用递归函数来走的树。使用跟踪,而不是警报()():

You need to use a recursive function to walk down the tree. Using trace() instead of alert():

private function urlLdr_complete(event:Event):void
{
    var xmlData:XML=new XML(URLLoader(event.currentTarget).data);
    showNodeName(xmlData);
}

private function showNodeName($node:XML):void
{
    // Trace the current node
    trace($node.name());
    if($node.hasChildNodes)
    {
        for each (var child:XML in $node.children())
        {
            // Recursively call this function on each child
            showNodeName(child);
        }
    }
}

或者,使用E4X的后裔()函数:

Or, use the E4X descendants() function:

private function urlLdr_complete(event:Event):void
{
    var xmlData:XML=new XML(URLLoader(event.currentTarget).data);
    // Trace the root node:
    trace(xmlData.name());
    // And trace all its descendants:
    for each(var child:XML in xmlData.descendants())
    {
        trace(child.name());
    }
}

这两个应该产生相同的结果:

Both should produce identical outcomes:

main
TabNavigator
NavigatorContent
NavigatorContent
TitleWindow
DataGrid
columns
ArrayList
GridColumn
GridColumn
GridColumn
Button

我没有测试过,但我希望内置的后裔()函数更有效率。

I haven't tested but I would expect the built-in descendants() function to be more efficient.